The general problem is this:
Given the Ksp, calculate the molar solubility (in mol/L) of a saturated solution of the substance.
Example #1: Silver chloride, AgCl, has a Ksp = 1.77 x 10¯10. Calculate its solubility in moles per liter.
1) Write the dissociation equation:
AgCl(s) ⇌ Ag+(aq) + Cl¯(aq)
2) Write the Ksp expression:
Ksp = [Ag+] [Cl¯]
This is the equation we must solve.
3) Put in the Ksp value:
1.77 x 10¯10 = [Ag+] [Cl¯]
4) We have to reason out the values of the two guys on the right. I do that by first assigning a variable, s, to the molar solubility of AgCl. That's the value that I want to determine.
To do this, I look at the molar relationship between AgCl and Ag+. (I do this because I want to express the concentrations of the ions by using 's.') I note that the molar relationship is a 1:1 molar ratio, meaning that for every one mole of AgCl that dissolves, one mole of Ag+ is produced. That leads me to this:[Ag+] = s
Now, I look at the relationship between AgCl and Cl¯. I see that it is also a 1:1 molar ratio, leading me to this:[Cl¯] = s
I am now ready to substitute into the Ksp expression.
5) Substituting, we get:
1.77 x 10¯10 = (s) (s)
7) Now, we take the square root of both sides. I hope I'm not too insulting when I emphasize both sides. I have had lots of people in my classes take the square root of the x2 side, but not the other. After the square root, we get:
s = 1.33 x 10¯5 M
This is the answer because there is a one-to-one relationship between the Ag+ dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯5 moles per liter.
One last thing. The Ksp value does not have any units on it, but when you get to the value for s, be sure to put M (for molarity) on it. The reasons behind this are complex and beyond the scope of the ChemTeam's goals for this web site.
Comment: it's important that you know how substances ionize. AgCl is easy, but you may not know that CuSCN (example #4) and silver azide (example #5) also ionize in a 1:1 molar ratio like AgCl.
Warning: You may know lots of common ions but, in a problem like the one under discussion, you may get an unusual one thrown at you on the test. Be prepared!
Example #2: Aluminum phosphate has a Ksp of 9.83 x 10¯21. What is its molar solubility in pure water?
1) Here is the dissociation equation:
AlPO4(s) ⇌ Al3+(aq) + PO43¯(aq)
2) Here is the Ksp expression:
Ksp = [Al3+] [PO43¯]
3) Keep in mind that the key point is the one-to-one ratio of the ions in solution. This means that the two ions are equal in their concentration. That allows this equation:
9.83 x 10¯21 = (s) (s)
4) Solving gives:
s = 9.91 x 10¯11 M
which is the answer.
Example #3: Calculate the molar solubility of barium sulfate, Ksp = 1.07 x 10¯10
1) The dissociation equation and the Ksp expression:
BaSO4(s) ⇌ Ba2+(aq) + SO42¯(aq)
Ksp = [Ba2+] [SO42¯]
2) The equation:
1.07 x 10¯10 = (s) (s)
3) when solved, gives:
s = 1.03 x 10¯5 M
Remember, this is the answer because the dissolved ions and the solid are also in a one-to-one molar ratio.
Notice how I did not say 'saturated solution' in the problem. When you see this, you need to assume that it is a saturated solution. Anything else makes the problem unworkable and that is not the intent of the question writer.
Example #4: CuSCN, Ksp = 1.77 x 10¯13
CuSCN(s) ⇌ Cu+(aq) + SCN¯(aq)
molar solubility = 4.21 x 10¯7 M
Example #5: Silver azide has the formula AgN3 and Ksp = 2.0 x 10¯8
AgN3(s) ⇌ Ag+(aq) + N3¯(aq)
Note azide, a fairly uncommon polyatomic ion. It could show up on the test! Calculating the molar solubility is left to the student.
Bonus Example: Magnesium ammonium phosphate, MgNH4PO4, Ksp = 2.5 x 10¯13
This is how the substance ionizes:
MgNH4PO4(s) ⇌ Mg2+(aq) + NH4PO42¯(aq)
The ammonium phosphate polyatomic ion, NH4PO42¯, is an uncommon one.